: chemistry qs: write the equillibrim constant for the reverse reaction

In summary: The correct value for K2 is 5x10^(-5).In summary, the equilibrium constant for the reverse reaction is 5x10^(-5).
  • #1
crosbykins
53
0
URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

Homework Statement



2. The equilibrium constant for the following reaction is 2.0 10^ 4

2HBr(g)  H2(g)  Br2(g)


Homework Equations



b) What is the equilibrium constant for the reverse reaction?


The Attempt at a Solution



I don't think this is right but

2.0 x 10^4 = (Br2)(H2)/(HBr)^2
2.0 x 10^4 = (x)(x)/(2x)^2
2.0 x 10^4 = 1/4

so for the reverse reaction K = 4 * 2.0 x 10^4.
 
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  • #2


crosbykins said:

I don't think this is right but

You're right, it's not :) but this is right:

crosbykins said:

2.0 x 10^4 = (Br2)(H2)/(HBr)^2

If that is the forward reaction, what would be the equation for the reverse reaction? (it might be easier to to just think of it as k = (A)(B)/(C) for C --> A + B )
 
  • #3


In equilibrium the forward and reverse reaction rates are the same.

Do you know what the equations for the reaction rates for each side are?
 
  • #4


p21bass said:
In equilibrium the forward and reverse reaction rates are the same.

Do you know what the equations for the reaction rates for each side are?

While it can be done through reaction rates - it is unnecessary complication. It can be done much faster and much easier just by writing the equation for the reverse reaction (see zhermes post).
 
  • #5


zhermes said:
You're right, it's not :) but this is right:


If that is the forward reaction, what would be the equation for the reverse reaction? (it might be easier to to just think of it as k = (A)(B)/(C) for C --> A + B )

ok it would be K = [HBr]^2/

[I2]

but how does that give the the value of K?

 
  • #6


Borek said:
While it can be done through reaction rates - it is unnecessary complication. It can be done much faster and much easier just by writing the equation for the reverse reaction (see zhermes post).

ok so the equation for the reverse reaction is 2HBr -> H2 + I2 and K = [HBr]^2/

[I2]

...how can i get the value of K from this

 
  • #7


[tex]K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4[/tex]

[tex]K_2 = \frac {[HBr]^2} {[H_2][Br_2]} [/tex]

If you still don't see the answer, try to calculate K1xK2.

Edit: note, that most of your equations are incorrect as they have HBr on one side of the reaction, I2 on the second.
 
  • #8


Borek said:
[tex]K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4[/tex]

[tex]K_2 = \frac {[HBr]^2} {[H_2][Br_2]} [/tex]

If you still don't see the answer, try to calculate K1xK2.

Edit: note, that most of your equations are incorrect as they have HBr on one side of the reaction, I2 on the second.

so K1 * K2 =1

that means K 2 must equal 1/2*10^ 4
 
Last edited:
  • #9


Wasn't that hard.
 
  • #10


crosbykins said:
so K1 * K2 =1

that means K 2 must equal 1/2*10^ 4

Not quite, try multiplying K1 and K2 to check.
 

What is an equilibrium constant?

The equilibrium constant, denoted by K, is a measure of the relative concentrations of the products and reactants at equilibrium for a chemical reaction. It helps to determine the direction in which a reaction will proceed.

How do you calculate the equilibrium constant for a reaction?

The equilibrium constant is calculated by dividing the concentration of the products by the concentration of the reactants, with each concentration raised to the power of its coefficient in the balanced chemical equation.

What is the significance of the equilibrium constant for the reverse reaction?

The equilibrium constant for the reverse reaction, denoted by K', is the reciprocal of the equilibrium constant for the forward reaction (K). It indicates how much the reverse reaction is favored compared to the forward reaction at equilibrium.

Can the equilibrium constant for the reverse reaction be greater than the equilibrium constant for the forward reaction?

Yes, it is possible for K' to be greater than K. This indicates that the reverse reaction is favored at equilibrium, and the reaction will proceed in the reverse direction.

How does temperature affect the equilibrium constant for the reverse reaction?

Changing the temperature can shift the equilibrium of a reaction, thus affecting the equilibrium constants. However, the ratio of K' to K remains the same regardless of the temperature.

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